About the Pendulum Period Calculator
This pendulum calculator uses the simple pendulum formula T = 2π√(L/g) to find how long one full swing takes, or works backwards to find the string length that gives a period you want — for example the 0.994 m “seconds pendulum” that ticks once per second each way. Choose Earth, the Moon, Mars, Jupiter or a custom gravity to see how the period changes on other worlds.
It is useful for physics homework and lab write-ups, clock and metronome projects, and anyone curious how swing length affects timing. Besides the period you get the frequency, angular frequency and, for the release angle you enter, the exact large-amplitude period and the speed of the bob at the bottom of its swing.
The simple formula assumes a point mass on a light, inextensible string with no air resistance and a small swing (under about 15°). The large-angle result uses the exact elliptic-integral solution, so you can see how much a wide swing slows the pendulum down.
How to use the pendulum period calculator
- 1Choose whether you know the length or the period.
- 2Enter the length (and its unit) or the period in seconds.
- 3Pick Earth or another gravity setting.
- 4Optionally enter the release angle for the exact large-swing period.
- 5Read the period or length, then compare with the chart.
Formula and method
For small swings a simple pendulum behaves like a harmonic oscillator, so its period depends only on the length L and gravitational acceleration g — not on the mass of the bob or (approximately) the swing size. Rearranging gives the length needed for a target period, L = gT² ÷ 4π². Frequency is the reciprocal of the period and angular frequency is ω = 2πf = √(g/L).
For larger amplitudes the exact period is T = 4√(L/g)·K(sin(θ₀/2)), where K is the complete elliptic integral of the first kind; the calculator evaluates it with the arithmetic-geometric mean. The bob’s speed at the lowest point comes from energy conservation: v = √(2gL(1 − cos θ₀)).
- T
- Period of one full swing (s)
- L
- Length from pivot to centre of mass (m)
- g
- Gravitational acceleration (m/s²)
- θ₀
- Release angle (amplitude)
- K
- Complete elliptic integral of the first kind
Worked examples
1 m pendulum on Earth
T = 2π√(1 ÷ 9.80665) = 2.006 s, a frequency of 0.498 Hz. Released at 10°, the exact period is 2.010 s — only 0.19% longer — and the bob passes the bottom at about 0.55 m/s.
Length for a 1-second period
L = 9.80665 × 1² ÷ (4π²) = 0.2484 m. A pendulum about 24.8 cm long completes a full back-and-forth swing every second.
The same 1 m pendulum on the Moon
With lunar gravity of 1.62 m/s², T = 2π√(1 ÷ 1.62) = 4.94 s — about 2.5 times slower than on Earth, because the period scales with 1/√g.
Frequently asked questions
What is the formula for the period of a pendulum?+
For small swings, T = 2π√(L/g), where L is the length from pivot to the bob’s centre of mass and g is gravitational acceleration (9.81 m/s² on Earth).
Does the mass of the bob affect the period?+
No. In the simple pendulum model the mass cancels out, so a heavy and a light bob on the same length of string swing with the same period (ignoring air resistance).
How long is a seconds pendulum?+
A seconds pendulum takes one second per one-way swing, a full period of 2 s. On Earth that needs a length of about 0.994 m.
Does the swing angle change the period?+
Slightly. At 10° the period is about 0.2% longer than the small-angle formula predicts, at 30° about 1.7% longer, and the difference grows quickly beyond that.
How does length affect a pendulum’s period?+
The period is proportional to the square root of the length, so making a pendulum four times longer doubles its period, and making it a quarter as long halves it.